Question

# Simple harmonic motion of a mass-on-a-vertical-spring is being demonstrated to a physics class. Students are asked...

Simple harmonic motion of a mass-on-a-vertical-spring is being demonstrated to a physics class. Students are asked to find the spring constant, k. After suspending a mass of 335.0 g from the spring, one student notices the spring is displaced 45.5 cm from its previous equilibrium. With this information, calculate the spring constant.

In the final section of the class, a student is asked to investigate the energy distribution of the spring system described above. The student pulls the mass down an additional 34.1 cm from the equilibrium point of 45.5 cm when the mass is stationary and allows the system to oscillate. Using the equilibrium point of 45.5 cm as the zero point for total potential energy, calculate the velocity and total potential energy for each displacement given and insert the correct answer using the \"choices\" column. (Scroll down to see the full table.)

according Hooke's law the increase in length in the spring is directly proportional to the force applied to it.

F is directly proportional to x
F=kx
Where k is the spring constant
Mass = 335 grams or 0.335 kg (converted grams to kilograms)

as we know F=mg
F=0.335*9.8
F=3.28 N

Now F=kx
3.28N=k * 0.455m (converted 45.5 cm into meter)
k=0.455/3.28
k=0.138 or 0.14 Nm

If the 29.5 cm
is a displacement beyond the new equilibrium position, then the potential energy is
(1/2)ky^2 = (1/2)(0.14 N/m)(0.335m)^2 = 0.00785 joules

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